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I am working to develop a deep reinforcement learning model for balancing an inverted double pendulum.

To do so, I needed to derive the equations of motion, via the Euler-Lagrange (EL) method and now need to linearise the EL equations to get the local stability near the equilibrium point, when:

$$\theta(t)=0\:\:\:\:\:\:\:\:\:\:\:\phi(t)=0$$

i.e., both pendulums are in the vertically upright position; vertically stable.

After reading several papers, the linearisation most commonly used, as described here is:

$$sin(\theta(t))\approx\theta(t)\:\:\:\:\:\:\:\:\:\:\:sin(\phi(t))\approx\phi(t)\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(1)$$

Using these linearisations, the final acceleration equations each have a term that is multiplied by either $\theta(t)$ or $\phi(t)$, which does not even make sense.

For example, one of the terms in the $x$-acceleraion equation is:

$$-M^{2} g l^{2} 4I + 4 M l^{2} - M l \theta(t)+...$$

As seen, the last term $M l \theta(t)$ is a multiplication of the mass, length and angle. How can a value just be multiplied by an angle?

Instead, if using, $\theta(t)=0$ & $\phi(t)=0$, then:

$$sin(\theta(t))=sin(0)=0\:\:\:\:\:\:\:\:\:\:\:sin(\phi(t))=sin(0)=0\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:(2)$$

Using the above linearisations, there are no values in the final acceleration equations that are being multiplied with just an angle?!?

Can someone please explain why the papers have used the linearisations stated in $(1)$ instead of $(2)$, or provide some links to the relevant literature? Intuitively, the linearisations of $(2)$ correlate with the initial idea of vertically stabilising the pendulums?

Thank you.

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1 Answer 1

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This answer comes late, but I hope it is still useful.

The second set of linearization linearize the problem to triviality. You are basically constraining the pendulae to not move. A more accurate terminology would be you are keeping them constant (by constraining), therefore eliminating the dynamics completely. If the system cannot move, then there is no balancing possible, even if the language might make you think otherwise, as balancing consists of making sure it doesn't move.

Notice that if $$\sin(\theta)=\sin(\phi)=0$$ then $$\theta=\phi=0 \pmod {2\pi}$$ and $$\dot{\theta} = \dot{\phi}=0 $$

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